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(1) n^2 - 1 is an odd integer --> \(n^2-1=odd\) --> \(n^2=odd+1=even\). Now, since \(n\) is an integer, then in order \(n^2\) to be even \(n\) must be even. Sufficient. Notice that if we were not told that \(n\) is an integer, then \(n\) could be some irrational number (square root of an even number), for example \(\sqrt{2}\), so not an even integer.

(2) 3n + 4 is an even integer --> \(3n + 4=even\) --> \(3n=even-4=even\). The same here, since \(n\) is an integer, then in order \(3n\) to be even \(n\) must be even. Sufficient. Notice that if we were not told that \(n\) is an integer, then \(n\) could be some fraction, for example \(\frac{2}{3}\), so not an even integer.

(1) n^2 - 1 is odd. The next consecutive integer is n^2 and is therefore even. This means that n must be even too, because squaring a number does NOT change this. --> sufficient

(2) 3n + 4 is even. So 3n is even, too. This means that the prime factorization of 3n includes at least one 2. Dividing by 3 (to get from 3n to n) does NOT eliminate recude the number of twos in the prime factorization, so n is even. --> sufficient

The correct answer is D. Both statements are individually sufficient.

i will go with d both options individually can get the required info -as its an integer no more fractions integer when squared gives the same type value i.e odd gives odd and even gives even same way multiplying odd with even gives even and odd with odd gives odd

hence both are individually sufficient
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(1) n^2 - 1 is an odd integer --> \(n^2-1=odd\) --> \(n^2=odd+1=even\). Now, since \(n\) is an integer, then in order \(n^2\) to be even \(n\) must be even. Sufficient. Notice that if we were not told that \(n\) is an integer, then \(n\) could be some irrational number (square root of an even number), for example \(\sqrt{2}\), so not an even integer.

(2) 3n + 4 is an even integer --> \(3n + 4=even\) --> \(3n=even-4=even\). The same here, since \(n\) is an integer, then in order \(3n\) to be even \(n\) must be even. Sufficient. Notice that if we were not told that \(n\) is an integer, then \(n\) could be some fraction, for example \(\frac{2}{3}\), so not an even integer.

(1) n^2 - 1 is an odd integer --> \(n^2-1=odd\) --> \(n^2=odd+1=even\). Now, since \(n\) is an integer, then in order \(n^2\) to be even \(n\) must be even. Sufficient. Notice that if we were not told that \(n\) is an integer, then \(n\) could be some irrational number (square root of an even number), for example \(\sqrt{2}\), so not an even integer.

(2) 3n + 4 is an even integer --> \(3n + 4=even\) --> \(3n=even-4=even\). The same here, since \(n\) is an integer, then in order \(3n\) to be even \(n\) must be even. Sufficient. Notice that if we were not told that \(n\) is an integer, then \(n\) could be some fraction, for example \(\frac{2}{3}\), so not an even integer.

(1) n^2 - 1 is an odd integer --> \(n^2-1=odd\) --> \(n^2=odd+1=even\). Now, since \(n\) is an integer, then in order \(n^2\) to be even \(n\) must be even. Sufficient. Notice that if we were not told that \(n\) is an integer, then \(n\) could be some irrational number (square root of an even number), for example \(\sqrt{2}\), so not an even integer.

(2) 3n + 4 is an even integer --> \(3n + 4=even\) --> \(3n=even-4=even\). The same here, since \(n\) is an integer, then in order \(3n\) to be even \(n\) must be even. Sufficient. Notice that if we were not told that \(n\) is an integer, then \(n\) could be some fraction, for example \(\frac{2}{3}\), so not an even integer.

Answer: D.

Since n is a integer, can we not try with n as 0?

Yes, n can be 0 but 0 is even too.
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(1) n^2 - 1 is an odd integer. (2) 3n + 4 is an even integer.

Given: n is an integer Required: is n even?

Statement 1: \(n^2\) - 1 is an odd integer \(n^2\) - 1 = (n-1)(n+1) = odd. This means both n-1 and n+1 are odd Odd*Odd = Odd Odd*Even = Even Even*Even = Even

n-1, n, n+1 are three consecutive integers. Since we know that both n-1 and n+1 are odd Hence n has to be even. SUFFICIENT

Statement 2: 3n + 4 is an even integer Even + Even = Even Even + Odd = Odd Odd + Odd + Odd

Since 3n+4 = even and 4 is an even integer. Hence 3n = even. Therefore n = even SUFFICIENT

(1) n^2 - 1 is an odd integer. (2) 3n + 4 is an even integer.

We need to determine whether integer n is even. Let's review four facts about even and odd integers: 1) An integer and its square are either both even or both odd. 2) The sum (or difference) between an even integer and an odd integer is always odd. 3) The sum of two even integers (or two odd integers) is always even. 4) If the product of two integers is even, at least one of them must be even.

Statement One Alone:

(n^2) - 1 is an odd integer.

Since (n^2) - 1 is an odd integer, we know that n^2 must be even and thus n must be even. Statement one is sufficient to answer the question. We can eliminate answer choices B, C, and E.

Statement Two Alone:

3n + 4 is an even integer.

Since 3n + even = even integer, we know that 3n must be even, and since 3 is odd, n must be even. Statement two is sufficient to answer the question.

The answer is D.
_________________

Jeffrey Miller Scott Woodbury-Stewart Founder and CEO

We need to find if x is even or odd given => x is an integer statement 1 =>n^2-1=odd => n^2=> even => n must be even Rule used => POWER Does not effect the even / odd nature of any number

statement 2 => 3n+4=even=> 3n=even => n must even Rulw => if XY=even => at-least one of them must be even Hence suff

Here is my solution -> Her we need to get the Even/Odd nature of integer n Lets dive into statements Statement 1 n^2-1=odd so n^2 must be even RULE-> POWER Does not affect the even/odd nature of any integer Hence n must be even too. Hence Sufficient Statement 2-> 3n+4=even Hence 3n=even-even=even As 3n=even and 3 is odd => n must be even to make 3n even Hence Sufficient Hence D _________________

Give me a hell yeah ...!!!!!

gmatclubot

Re: If n is an integer, is n even?
[#permalink]
03 Dec 2016, 17:23

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